Question 1034214
{{{((4x^(-3)y^2)/(xy^(-5)))^(-2)=(4x^(-3-1)y^(2-(-5)))^(-2)}}}
{{{((4x^(-3)y^2)/(xy^(-5)))^(-2)=(4x^(-4)y^(7))^(-2)}}}
{{{((4x^(-3)y^2)/(xy^(-5)))^(-2)=(1/16)x^(8)y^(-14)}}}
{{{((4x^(-3)y^2)/(xy^(-5)))^(-2)=highlight(x^(8)/(16y^4))}}}